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Does the elasticity of a collision depend on the object's mass?

Physics Asked on January 2, 2022

This refers to my other question Why is the ideal gas law only valid for hydrogen?. In the update, my teacher said that hydrogen is closer to an ideal gas because its mass is lower: $m_{rm H} thickapprox displaystylefrac {1}4 m_{rm He}$. Since the mass of the object is not included in the ideal Gas law which is $PV=nRT$, I concluded that they probably mean that the mass is relevant to the elasticity of a collision (which is one of the properties of an ideal gas).

Is that true? I see that the mass of the objects seems to play a role as it is included in the formula for elastic collisions:

$$u=frac {m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

But this equation does not describe how elastic a collision is, it only describes the behaviour of two objects after a perfectly elastic collision.

As far as I know, the elasticity of a collision is mainly determined by the ability of the objects to deform non-permanently (e.g. billiard balls or bouncy balls are close to being elastic on the macroscopic scale).

In my example, I compare the elasticity of the collision between two hydrogen or rather helium atoms. Does the higher mass of helium (and thus the higher volume of its nucleus) affect the elasticity of the collision?

As explained in the other question, I would be grateful if you could provide some sources for your answers in the case they cannot be simply deduced by known formulas or facts.

One Answer

In the update, my teacher said that hydrogen is closer to an ideal gas because its mass is lower: $M_{H}$≈1/4 $m_{He}$

As pointed out by @nasu the mass of the hydrogen gas molecule is 1/2 of helium, not 1/4. You are comparing the masses of the atoms.

Although the mass of the hydrogen gas molecule is less than helium, the radius of the hydrogen atom, 53pm, is greater than the helium atom, 31pm. So the size of the helium gas atom is less then the hydrogen diatomic molecule. A gas behaves more ideally the smaller its size relative to the separation between atoms/molecules, all other things being equal.

In addition, all other things equal, there will be fewer collisions between the helium atoms than between the hydrogen molecules. In this regard, helium has a smaller "kinetic diameter" (260pm) than hydrogen (289pm). Per Wikipedia the "kinetic diameter is a measure applied to atoms and molecules that expresses the likelihood that a molecule in a gas will collide with another molecule. It is an indication of the size of the molecule as a target."

Since the mass of the object is not included in the ideal Gas law which is $PV=nRT$, I concluded that they probably mean that the mass is relevant to the elasticity of a collision (which is one of the properties of an ideal gas).

The mass is included in the equation since the number of moles $n$ of the gas is the mass of the gas divided by its molecular weight. The ideal gas law can also be written in terms of mass $m$ as:

$$PV=mR_{g}T$$

Where in this case $R_g$ is the specific gas constant (specific to the gas under consideration). $R$ in the formula $PV=nRT$ is the universal gas constant.

Is that true? I see that the mass of the objects seems to play a role as it is included in the formula for elastic collisions:

$$u=frac {m_1 v_1 + m_2 v_2}{m_1 + m_2}$$

But this equation does not describe how elastic a collision is, it only describes the behaviour of two objects after a perfectly elastic collision.

You're equation does not appear to be for an elastic collision. It seems to be for a perfectly inelastic collision where the two objects stick together following the collision with a final velocity of $u$, based on conservation of momentum. In any case, I have not heard of mass, as a fundamental property of matter, playing a role in the elasticity of the collision. But I would be interested if anyone else has knowledge to the contrary.

In my example, I compare the elasticity of the collision between two hydrogen or rather helium atoms. Does the higher mass of helium (and thus the higher volume of its nucleus) affect the elasticity of the collision?

Again, I don't see how a higher mass affects the elasticity of the collision. I only see that it affects the final momenta and kinetic energies of the colliding objects. To the best of my knowledge, it is the mechanical properties of the materials (are they elastic? Viscoelastic? (part elastic and part inelastic), etc.) that determine collision elasticity. Again, however, perhaps someone else can point to a reliable source to the contrary.

I am pretty sure that this collision is for elastic collisions. At least, this is what I also found on Wikipedia.

I looked at the Wikipedia article. It looks like $u$ in your equation is the velocity in the center of mass frame which does not change before and after the collision. Since you didn't state was $u$ was I assumed it was the velocity of the two masses stuck together following the collision, which would also satisfy conservation of momentum.

Regardless, I believe your equation only requires conservation of momentum. It would apply whether the collision is elastic or inelastic. So the equation does not only apply to elastic collisions, as stated.

Hope this helps.

Answered by Bob D on January 2, 2022

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